What is $f (x) = \int x - sin 2 x + cos x d x$ $f(x) = int x-sin2x+cosx dx$ if $f (\frac{π}{2}) = 3$ $f(pi/2)=3$ ?

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What is $f (x) = \int x - sin 2 x + cos x d x$ $f(x) = int x-sin2x+cosx dx$ if $f (\frac{π}{2}) = 3$ $f(pi/2)=3$ ?

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Answer 1 · 2017-02-04T19:27:13

$f (x) = \frac{x^{2}}{2} + \frac{1}{2} cos (2 x) + sin (x) + \frac{5}{2} - \frac{π^{2}}{8}$ $f(x)=x^2/2+1/2cos(2x)+sin(x)+5/2-pi^2/8$

Explanation:

We will use the following rules:

$\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x$ $intf(x)+-g(x)dx=intf(x)dx+-intg(x)dx$
$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$
$\int sin (u) d u = - cos (u) + C$ $intsin(u)du=-cos(u)+C$
$\int cos (u) d u = sin (u) + C$ $intcos(u)du=sin(u)+C$

So:

$f (x) = \int x^{1} d x - \int sin (2 x) d x + \int cos (x) d x$ $f(x)=intx^1dx-intsin(2x)dx+intcos(x)dx$

The first and third can be found directly:

$f (x) = \frac{x^{2}}{2} - \int sin (2 x) d x + sin (x)$ $f(x)=x^2/2-intsin(2x)dx+sin(x)$

For the remaining integral, let $u = 2 x$ $u=2x$ so $d u = (2) d x$ $du=(2)dx$ . Then:

$f (x) = \frac{x^{2}}{2} - \frac{1}{2} \int sin (2 x) (2) d x + sin (x)$ $f(x)=x^2/2-1/2intsin(2x)(2)dx+sin(x)$

$f (x) = \frac{x^{2}}{2} - \frac{1}{2} \int sin (u) d u + sin (x)$ $f(x)=x^2/2-1/2intsin(u)du+sin(x)$

$f (x) = \frac{x^{2}}{2} - \frac{1}{2} (- cos (u)) + sin (x) + C$ $f(x)=x^2/2-1/2(-cos(u))+sin(x)+C$

$f (x) = \frac{x^{2}}{2} + \frac{1}{2} cos (2 x) + sin (x) + C$ $f(x)=x^2/2+1/2cos(2x)+sin(x)+C$

We can now solve for $C$ $C$ using $f (\frac{π}{2}) = 3$ $f(pi/2)=3$ :

$3 = \frac{{(\frac{π}{2})}^{2}}{2} + \frac{1}{2} cos (2 \cdot \frac{π}{2}) + sin (\frac{π}{2}) + C$ $3=(pi/2)^2/2+1/2cos(2*pi/2)+sin(pi/2)+C$

$3 = \frac{π^{2}}{8} + \frac{1}{2} (- 1) + 1 + C$ $3=pi^2/8+1/2(-1)+1+C$

$C = \frac{5}{2} - \frac{π^{2}}{8}$ $C=5/2-pi^2/8$

So:

$f (x) = \frac{x^{2}}{2} + \frac{1}{2} cos (2 x) + sin (x) + \frac{5}{2} - \frac{π^{2}}{8}$ $f(x)=x^2/2+1/2cos(2x)+sin(x)+5/2-pi^2/8$

Answer 2 · 2017-02-04T19:46:15

$f (x) = \frac{x^{2}}{2} + \frac{1}{2} cos 2 x + sin x + \frac{20 - π^{2}}{8}$ $f(x)=x^2/2+1/2cos2x+sinx+(20-pi^2)/8$

Explanation:

$f (x) = \int x - sin 2 x + cos x d x$ $f(x)=intx-sin2x+cosxdx$
$" "$
$f (x) = \int x d x - \int sin 2 x d x + \int cos x d x + C$ $f(x)=intxdx-intsin2xdx+intcosxdx+C$ $C$ $" "C$ is a constant
$" "$
$f (x) = \frac{x^{2}}{2} - \frac{1}{2} \int d (- cos 2 x) + \int d (sin x) + C$ $f(x)=x^2/2-1/2intd(-cos2x)+intd(sinx)+C$
$" "$
$f (x) = \frac{x^{2}}{2} + \frac{1}{2} cos 2 x + sin x + C$ $f(x)=x^2/2+1/2cos2x+sinx+C$
$" "$
$" "$
Le us find $C$ $C$
$" "$
Given $f (\frac{π}{2}) = 3$ $f(pi/2)=3$
$" "$
$f (\frac{π}{2}) = \frac{{(\frac{π}{2})}^{2}}{2} + \frac{1}{2} cos (2 \cdot \frac{π}{2}) + sin (\frac{π}{2}) + C$ $f(pi/2)=(pi/2)^2/2+1/2cos(2*pi/2)+sin(pi/2)+C$
$" "$
Substituting $f (\frac{π}{2}) = 3$ $f(pi/2)=3$
$" "$
$\Rightarrow 3 = \frac{{(π)}^{2}}{8} + \frac{1}{2} cos π + 1 + C$ $rArr3=(pi)^2/8+1/2cospi+1+C$
$" "$
$\Rightarrow 3 = \frac{{(π)}^{2}}{8} - \frac{1}{2} + 1 + C$ $rArr3=(pi)^2/8-1/2+1+C$
$" "$
$\Rightarrow 3 = \frac{π^{2}}{8} + \frac{1}{2} + C$ $rArr3=pi^2/8+1/2+C$
$" "$
$\Rightarrow C = 3 - \frac{1}{2} - \frac{π^{2}}{8}$ $rArrC=3-1/2-pi^2/8$
$" "$
$\Rightarrow C = \frac{24}{8} - \frac{4}{8} - \frac{π^{2}}{8}$ $rArrC=24/8-4/8-pi^2/8$
$" "$
$\Rightarrow C = \frac{20 - π^{2}}{8}$ $rArrC=(20-pi^2)/8$
$" "$
$" "$
Therefore, $f (x) = \frac{x^{2}}{2} + \frac{1}{2} cos 2 x + sin x + \frac{20 - π^{2}}{8}$ $f(x)=x^2/2+1/2cos2x+sinx+(20-pi^2)/8$

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What is $f (x) = \int x - sin 2 x + cos x d x$ $f(x) = int x-sin2x+cosx dx$ if $f (\frac{π}{2}) = 3$ $f(pi/2)=3$ ?

2 Answers

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